Question

Air undergoes dielectric breakdown at a field strength of 3 MV/m. Could you store energy in an electric field in air with the same energy density as gasoline? The energy content of gasoline 44* 10^6 J/kg and the density of gasoline is 670 kg/m^3. Yes or No.

Answer 1

Answer:

No

Explanation:

The energy content of gasoline is

$44\cdot 10^6 J/kg$

while the density is

$670 kg/m^3$

So the energy density of gasoline is

$u = (44\cdot 10^6 J/kg)(670 kg/m^3)=2.95\cdot 10^{10} J/m^3$

The energy density of an electric field is given by

$u_E = \frac{1}{2}\epsilon_0 E^2$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

E is the strength of the electric field

For air at dielectric breakdown,

$E=3 MV/m = 3\cdot 10^6 V/m$

Substituting into the equation,

$u_E = \frac{1}{2}(8.85\cdot 10^{-12})(3\cdot 10^6)^2=39.8 J/m^3$

We see that $u_E < u$, so the energy density of the electric field is much lower than the energy content of gasoline, so the answer is No.